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CyC2018
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@ -35,17 +35,21 @@ Java 的 I/O 大概可以分成以下几类：
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# 二、磁盘操作
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File 类可以用于表示文件和目录，但是它只用于表示文件的信息，而不表示文件的内容。
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File 类可以用于表示文件和目录的信息，但是它不表示文件的内容。
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递归地输出一个目录下所有文件：
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```java
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public static void listAllFiles(File dir) {
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public static void listAllFiles(File dir)
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{
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if (dir == null || !dir.exists()) {
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return;
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}
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if (dir.isFile()) {
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System.out.println(dir.getName());
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return;
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}
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for (File file : dir.listFiles()) {
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for (File file : Objects.requireNonNull(dir.listFiles())) {
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listAllFiles(file);
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}
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}
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@ -56,12 +60,13 @@ public static void listAllFiles(File dir) {
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使用字节流操作进行文件复制：
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```java
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public static void copyFile(String src, String dist) throws IOException {
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FileInputStream in = new FileInputStream("file/1.txt");
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FileOutputStream out = new FileOutputStream("file/2.txt");
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public static void copyFile(String src, String dist) throws IOException
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{
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FileInputStream in = new FileInputStream(src);
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FileOutputStream out = new FileOutputStream(dist);
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byte[] buffer = new byte[20 * 1024];
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// read() 最多读取 buffer.length 个字节，返回的是实际读取的个数
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// 返回 -1 的时候表示读到 eof，即文件尾
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/* read() 最多读取 buffer.length 个字节
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返回的是实际读取的个数，返回 -1 的时候表示读到 eof，即文件尾 */
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while (in.read(buffer, 0, buffer.length) != -1) {
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out.write(buffer);
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}
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@ -87,7 +92,8 @@ DataInputStream 装饰者提供了对更多数据类型进行输入的操作，
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不管是磁盘还是网络传输，最小的存储单元都是字节，而不是字符。但是在程序中操作的通常是字符形式的数据，因此需要提供对字符进行操作的方法。
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InputStreamReader 实现从文本文件的字节流解码成字符流；OutputStreamWriter 实现字符流编码成为文本文件的字节流。
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- InputStreamReader 实现从文本文件的字节流解码成字符流；
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- OutputStreamWriter 实现字符流编码成为文本文件的字节流。
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逐行输出文本文件的内容：
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@ -98,8 +104,9 @@ String line;
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while ((line = bufferedReader.readLine()) != null) {
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System.out.println(line);
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}
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// 装饰者模式使得 BufferedReader 组合了一个 Reader 对象
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// 在调用 BufferedReader 的 close() 方法时会去调用 fileReader 的 close() 方法，因此只要一个 close() 调用即可
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/* 装饰者模式使得 BufferedReader 组合了一个 Reader 对象
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在调用 BufferedReader 的 close() 方法时会去调用 fileReader 的 close() 方法
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因此只要一个 close() 调用即可 */
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bufferedReader.close();
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```
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@ -140,31 +147,35 @@ byte[] bytes = str1.getBytes();
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序列化的类需要实现 Serializable 接口，它只是一个标准，没有任何方法需要实现，但是如果不去实现它的话而进行序列化，会抛出异常。
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```java
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public static void main(String[] args) throws IOException, ClassNotFoundException {
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public static void main(String[] args) throws IOException, ClassNotFoundException
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{
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A a1 = new A(123, "abc");
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String objectFile = "file/a1";
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ObjectOutputStream objectOutputStream = new ObjectOutputStream(new FileOutputStream(objectFile));
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objectOutputStream.writeObject(a1);
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objectOutputStream.close();
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ObjectInputStream objectInputStream = new ObjectInputStream(new FileInputStream(objectFile));
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A a2 = (A) objectInputStream.readObject();
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objectInputStream.close();
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System.out.println(a2);
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}
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private static class A implements Serializable {
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private static class A implements Serializable
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{
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private int x;
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private String y;
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A(int x, String y) {
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A(int x, String y)
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{
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this.x = x;
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this.y = y;
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}
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@Override
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public String toString() {
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return "x = " + x +
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" " + "y = " + y;
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public String toString()
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{
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return "x = " + x + " " + "y = " + y;
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}
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}
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```
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@ -202,16 +213,19 @@ InetAddress.getByAddress(byte[] address);
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可以直接从 URL 中读取字节流数据。
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```java
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URL url = new URL("http://www.baidu.com");
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InputStream is = url.openStream(); // 字节流
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InputStreamReader isr = new InputStreamReader(is, "utf-8"); // 字符流
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BufferedReader br = new BufferedReader(isr);
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String line = br.readLine();
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while (line != null) {
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System.out.println(line);
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line = br.readLine();
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public static void main(String[] args) throws IOException
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{
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URL url = new URL("http://www.baidu.com");
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InputStream is = url.openStream(); /* 字节流 */
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InputStreamReader isr = new InputStreamReader(is, "utf-8"); /* 字符流 */
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BufferedReader br = new BufferedReader(isr);
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String line = br.readLine();
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while (line != null) {
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System.out.println(line);
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line = br.readLine();
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}
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br.close();
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}
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br.close();
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```
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## Sockets
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@ -2526,7 +2526,8 @@ public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum) {
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正确的解法应该是和书上一样，先旋转每个单词，再旋转整个字符串。
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```java
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public String ReverseSentence(String str) {
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public String ReverseSentence(String str)
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{
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int n = str.length();
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char[] chars = str.toCharArray();
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int i = 0, j = 0;
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@ -2541,12 +2542,14 @@ public String ReverseSentence(String str) {
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return new String(chars);
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}
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private void reverse(char[] c, int i, int j) {
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private void reverse(char[] c, int i, int j)
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{
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while (i < j)
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swap(c, i++, j--);
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}
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private void swap(char[] c, int i, int j) {
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private void swap(char[] c, int i, int j)
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{
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char t = c[i];
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c[i] = c[j];
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c[j] = t;
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@ -2566,7 +2569,8 @@ private void swap(char[] c, int i, int j) {
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将 "abcXYZdef" 旋转左移三位，可以先将 "abc" 和 "XYZdef" 分别旋转，得到 "cbafedZYX"，然后再把整个字符串旋转得到 "XYZdefabc"。
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```java
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public String LeftRotateString(String str, int n) {
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public String LeftRotateString(String str, int n)
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{
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if (n >= str.length())
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return str;
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char[] chars = str.toCharArray();
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@ -2576,12 +2580,14 @@ public String LeftRotateString(String str, int n) {
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return new String(chars);
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}
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private void reverse(char[] chars, int i, int j) {
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private void reverse(char[] chars, int i, int j)
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{
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while (i < j)
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swap(chars, i++, j--);
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}
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private void swap(char[] chars, int i, int j) {
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private void swap(char[] chars, int i, int j)
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{
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char t = chars[i];
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chars[i] = chars[j];
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chars[j] = t;
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@ -2599,15 +2605,16 @@ private void swap(char[] chars, int i, int j) {
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## 解题思路
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```java
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public ArrayList<Integer> maxInWindows(int[] num, int size) {
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public ArrayList<Integer> maxInWindows(int[] num, int size)
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{
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ArrayList<Integer> ret = new ArrayList<>();
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PriorityQueue<Integer> heap = new PriorityQueue<Integer>((o1, o2) -> o2 - o1);
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if (size > num.length || size < 1)
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return ret;
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PriorityQueue<Integer> heap = new PriorityQueue<>((o1, o2) -> o2 - o1); /* 大顶堆 */
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for (int i = 0; i < size; i++)
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heap.add(num[i]);
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ret.add(heap.peek());
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for (int i = 1, j = i + size - 1; j < num.length; i++, j++) {
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for (int i = 1, j = i + size - 1; j < num.length; i++, j++) { /* 维护一个大小为 size 的大顶堆 */
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heap.remove(num[i - 1]);
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heap.add(num[j]);
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ret.add(heap.peek());
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@ -2633,15 +2640,17 @@ public ArrayList<Integer> maxInWindows(int[] num, int size) {
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空间复杂度：O(N<sup>2</sup>)
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```java
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public List<Map.Entry<Integer, Double>> dicesSum(int n) {
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public List<Map.Entry<Integer, Double>> dicesSum(int n)
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{
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final int face = 6;
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final int pointNum = face * n;
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long[][] dp = new long[n + 1][pointNum + 1];
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for (int i = 1; i <= face; i++)
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dp[1][i] = 1;
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for (int i = 2; i <= n; i++)
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for (int j = i; j <= pointNum; j++) // 使用 i 个骰子最小点数为 i
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for (int j = i; j <= pointNum; j++) /* 使用 i 个骰子最小点数为 i */
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for (int k = 1; k <= face && k <= j; k++)
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dp[i][j] += dp[i - 1][j - k];
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@ -2649,6 +2658,7 @@ public List<Map.Entry<Integer, Double>> dicesSum(int n) {
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List<Map.Entry<Integer, Double>> ret = new ArrayList<>();
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for (int i = n; i <= pointNum; i++)
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ret.add(new AbstractMap.SimpleEntry<>(i, dp[n][i] / totalNum));
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return ret;
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}
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```
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@ -2658,24 +2668,30 @@ public List<Map.Entry<Integer, Double>> dicesSum(int n) {
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空间复杂度：O(N)
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```java
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public List<Map.Entry<Integer, Double>> dicesSum(int n) {
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public List<Map.Entry<Integer, Double>> dicesSum(int n)
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{
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final int face = 6;
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final int pointNum = face * n;
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long[][] dp = new long[2][pointNum + 1];
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for (int i = 1; i <= face; i++)
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dp[0][i] = 1;
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int flag = 1;
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int flag = 1; /* 旋转标记 */
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for (int i = 2; i <= n; i++, flag = 1 - flag) {
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for (int j = 0; j <= pointNum; j++)
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dp[flag][j] = 0; // 旋转数组清零
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for (int j = i; j <= pointNum; j++) // 使用 i 个骰子最小点数为 i
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dp[flag][j] = 0; /* 旋转数组清零 */
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for (int j = i; j <= pointNum; j++)
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for (int k = 1; k <= face && k <= j; k++)
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dp[flag][j] += dp[1 - flag][j - k];
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}
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final double totalNum = Math.pow(6, n);
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List<Map.Entry<Integer, Double>> ret = new ArrayList<>();
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for (int i = n; i <= pointNum; i++)
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ret.add(new AbstractMap.SimpleEntry<>(i, dp[1 - flag][i] / totalNum));
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return ret;
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}
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```
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@ -2691,18 +2707,20 @@ public List<Map.Entry<Integer, Double>> dicesSum(int n) {
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## 解题思路
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```java
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public boolean isContinuous(int[] nums) {
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public boolean isContinuous(int[] nums)
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{
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if (nums.length < 5)
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return false;
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Arrays.sort(nums);
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int cnt = 0;
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for (int num : nums)
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for (int num : nums) /* 统计癞子数量 */
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if (num == 0)
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cnt++;
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for (int i = cnt; i < nums.length - 1; i++) {
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if (nums[i + 1] == nums[i])
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return false;
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cnt -= nums[i + 1] - nums[i] - 1;
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cnt -= nums[i + 1] - nums[i] - 1; /* 使用癞子去补全不连续的顺子 */
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}
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return cnt >= 0;
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}
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@ -2721,10 +2739,11 @@ public boolean isContinuous(int[] nums) {
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约瑟夫环，圆圈长度为 n 的解可以看成长度为 n-1 的解再加上报数的长度 m。因为是圆圈，所以最后需要对 n 取余。
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```java
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public int LastRemaining_Solution(int n, int m) {
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if (n == 0)
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public int LastRemaining_Solution(int n, int m)
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{
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if (n == 0) /* 特殊输入的处理 */
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return -1;
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if (n == 1)
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if (n == 1) /* 返回条件 */
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return 0;
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return (LastRemaining_Solution(n - 1, m) + m) % n;
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}
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@ -2743,13 +2762,13 @@ public int LastRemaining_Solution(int n, int m) {
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使用贪心策略，假设第 i 轮进行卖出操作，买入操作价格应该在 i 之前并且价格最低。
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```java
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public int maxProfit(int[] prices) {
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public int maxProfit(int[] prices)
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{
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if (prices == null || prices.length == 0)
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return 0;
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int n = prices.length;
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int soFarMin = prices[0];
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int maxProfit = 0;
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for (int i = 1; i < n; i++) {
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for (int i = 1; i < prices.length; i++) {
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soFarMin = Math.min(soFarMin, prices[i]);
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maxProfit = Math.max(maxProfit, prices[i] - soFarMin);
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}
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@ -2774,7 +2793,8 @@ public int maxProfit(int[] prices) {
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以下实现中，递归的返回条件为 n <= 0，取非后就是 n > 0，递归的主体部分为 sum += Sum_Solution(n - 1)，转换为条件语句后就是 (sum += Sum_Solution(n - 1)) > 0。
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```java
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public int Sum_Solution(int n) {
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public int Sum_Solution(int n)
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{
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int sum = n;
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boolean b = (n > 0) && ((sum += Sum_Solution(n - 1)) > 0);
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return sum;
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@ -2796,8 +2816,9 @@ a ^ b 表示没有考虑进位的情况下两数的和，(a & b) << 1 就是进
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递归会终止的原因是 (a & b) << 1 最右边会多一个 0，那么继续递归，进位最右边的 0 会慢慢增多，最后进位会变为 0，递归终止。
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```java
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public int Add(int num1, int num2) {
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return num2 == 0 ? num1 : Add(num1 ^ num2, (num1 & num2) << 1);
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public int Add(int a, int b)
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{
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return b == 0 ? a : Add(a ^ b, (a & b) << 1);
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}
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```
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@ -2812,12 +2833,13 @@ public int Add(int num1, int num2) {
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## 解题思路
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```java
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public int[] multiply(int[] A) {
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public int[] multiply(int[] A)
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{
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int n = A.length;
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int[] B = new int[n];
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for (int i = 0, product = 1; i < n; product *= A[i], i++)
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for (int i = 0, product = 1; i < n; product *= A[i], i++) /* 从左往右累乘 */
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B[i] = product;
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for (int i = n - 1, product = 1; i >= 0; product *= A[i], i--)
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for (int i = n - 1, product = 1; i >= 0; product *= A[i], i--) /* 从右往左累乘 */
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B[i] *= product;
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return B;
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}
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@ -2844,17 +2866,18 @@ Output:
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## 解题思路
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```java
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public int StrToInt(String str) {
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public int StrToInt(String str)
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{
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if (str == null || str.length() == 0)
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return 0;
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boolean isNegative = str.charAt(0) == '-';
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int ret = 0;
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for (int i = 0; i < str.length(); i++) {
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char c = str.charAt(i);
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if (i == 0 && (c == '+' || c == '-'))
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if (i == 0 && (c == '+' || c == '-')) /* 符号判定 */
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continue;
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if (c < '0' || c > '9')
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return 0; // 非法输入
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if (c < '0' || c > '9') /* 非法输入 */
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return 0;
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ret = ret * 10 + (c - '0');
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}
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return isNegative ? -ret : ret;
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@ -2891,7 +2914,7 @@ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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[Leetcode : 236. Lowest Common Ancestor of a Binary Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/)
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在左右子树中查找两个节点的最低公共祖先，如果在其中一颗子树中查找到，那么就返回这个解，否则可以认为根节点就是最低公共祖先。
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在左右子树中查找是否存在 p 或者 q，如果 p 和 q 分别在两个子树中，那么就说明根节点就是 LCA。
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```java
|
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public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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Loading…
Reference in New Issue
Block a user